(cosA)^2+(cosB)^2+(cosC)^2=1

Question

A,B,C都是锐角,
求证:(tanAtanBtanC)^2大于等于8

Answer 1

移项并利用均值不等式得;
(sinA)^2=1-(cosA)^2=(cosB)^2+(cosC)^2>=2cosBcosC
同理
(sinB)^2>=2cosAcosC
(sinC)^2>=2cosAcosB
因为A,B,C都是锐角，其三角函数值都为正数.
因此可以把上面三式相乘，得
(sinAsinBsinC)^2>=8(cosAcosBcosC)^2
两边除以(cosAcosBcosC)^2即得
(tanAtanBtanC)^2>=8
取等号当且仅当
A=B=C=arccos(1/sqrt(3)) {或者cosA=cosB=cosC=1/根号3}

Answer 2

1/2(1+cos2a)+1/2(1+cos2b)+cosc^2=1
1/2(cos2a+cos2b)+cosc^2=0
cos(a+b)cos(a-b)+cosc^2=0
-cosccos(a-b)+cosc^2=0
cosc=0 cos(a-b)=cosc
c=90 c=a-b也就是a＝b+c a＝90
tanc无意义或 tana无意义